how many ways are there to collect $\$24$ from $4$ children and $6$ adults if each person gives at least $\$1$, but each child can give at most $\$4$ and each adult at most $\$7$?
answer:
$$(x+x^2 + x^3 + x^4)^4 (x+x^2 + \cdots + x^7)^6 = x^{10}(1+x + x^2 + x^3)^4 (1+x+x^2 + \cdots + x^6)^4$$
Using the argument: $\dfrac{1-x^{n-1}}{1-x}=1+x+x^2+x^3+\cdots+x^n.$
Note that: $$x^{10}(1+x + x^2 + x^3)^4 (1+x+x^2 + \cdots + x^6)^6 = x^{10}\left(\dfrac{1-x^4}{1-x}\right)^4\left(\dfrac{1-x^7}{1-x}\right)^6 = \dfrac{x^{10}\cdot(1-x^4)^4 \cdot(1-x^7)^6}{(1-x)^{10}}$$
Answer: $$(1-x^4)^4=1-4x^4+6x^8-4x^{12}+x^{16}$$ and $$(1-x^7)^6 = 1 - 6x^7+15x^4-20x^{21}+15x^{28}-6x^{35}+x^{42} $$
Thus: $${{14+9}\choose{9}}+(-6){{7+9}\choose{9}}+15{{0+9}\choose{9}}+(-4){{10+9}\choose{9}}+(-4)(-6){{3+9}\choose{9}}+6{{6+9}\choose{9}}+(-4){{2+9}\choose{9}}=414143$$
Is that right?